Question

Evaluate : $\int\limits^{\pi/2}_{0} {\cos^{2}x} \, dx$

Answer 1

Solution:

As we know that

${cos}^{2} x = \frac{1 + cos \: 2x}{2}$
put this value in the given function

$\int\limits^{\pi/2}_{0} {\cos^{2}x} \, dx = \int\limits^{\pi/2}_{0} { \frac{1 + cos \: 2x}{2} } \, dx \\ \\ = \int\limits^{\pi/2}_{0}( \frac{1}{2} + \frac{cos \: 2x}{2} )dx \\ \\ = [\frac{x}{2} + \frac{sin \: 2x}{4}]$

now put upper and lower limits

$= \frac{ {\pi} }{2} - 0 + \frac{sin \: \pi}{4} - sin \: 0 \\ \\ = \frac{\pi}{2} - 0 + \frac{0}{4} - 0 \\ \\ \int\limits^{\pi/2}_{0} {\cos^{2}x} \, dx= \frac{\pi}{2}$